Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a=1/1x2+1/2x3+....+1/99x100
a=1-1/2+1/2-1/3+....+1/99-1/100
a=1-1/100
a=99/100
b=4/1x3+4/3x5+.....+4/51x53
b=2x(2/1x3+2/3x5+....+2/51x53)
b=2x(1-1/3+1/3-1/5+...+1/51-1/53)
b=2x(1-1/53)
b=2x52/53
b=104/53
đúng tick cho mình nha
B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101
B= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 + ... + 1/99 - 1/101
B= 1/3 - 1/101
B=98/303
( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow M=1-\frac{1}{100}\)
\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)
\(a,M=1-\frac{1}{100}=\frac{99}{100}\)
\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(=1-\frac{1}{99}=\frac{98}{99}\)
=>\(N=\frac{98}{99}:2=\frac{49}{99}\)
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
\(a,\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
a)=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/99-1/100)
=1-1/2+1/2-1/3+1/3-1/4+......+1/99-1/100
=1-1/100=99/100
Bài 1:
A = \(\dfrac{1}{1\times3}\) + \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) +...+ \(\dfrac{1}{2019\times2021}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{1\times3}\) + \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+...+ \(\dfrac{2}{2019\times2021}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+...+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1010}{2021}\)